1
Integrate \(\int 4e^{-7x}dx\) using substitution:
Answer:
\(4\int e^{-7x}dx\) then for \(e^{-7x}\), substitute \(u=-7x\) and \(du=-7dx\) to get: \(-\frac{4}{7}\int e^{u}du=-\frac{4}{7}e^{u}+c\) then subbing back in for \(u=-7x\) I get: \(\int 4e^{-7x}dx=\boxed{-\frac{4}{7}e^{-7x}+c}\)
2
Biologists are treating a pond contaminated with bacteria. The level of contamination is changing at a rate of \(\frac{dN}{dt} =\frac{3150}{t^4}-220\) bacteria per cubic centimeter per day, where \(t\) is the number of days since treatment began. Find a function \(N(t)\) to estimate the level of contamination if the level after \(1\) day was \(6530\) bacteria per cubic centimeter.
Answer:
\(\frac{d}{dt}N = \frac{3150}{t^4}-220 \rightarrow dN = \big[\frac{3150}{t^4}-220 \big]dt \implies N = \int \frac{3150}{t^4}dt-\int 220dt\). Since \(N(1) = 6530 = N_o\), this leads to \(N = N_o -\frac{3150}{3t^3}-220t \implies N(1) = N_o - \frac{3150}{(1)^3}-220(1)\).
To get \(N_o\), I evaluate: \(6530+1050+220 = 7800= N_o\). This provides the function requested: \(\boxed{N(t)=7800-\frac{1050}{t^3}-220t}\)
3
Find the total area of the red rectangles in the figure below, where the equation of the line is \(f(x) = 2x-9\).
Answer:
By inspection: \(\boxed{16}\). By direct calcuation: \(\int_{4.5}^{8.5} f(x)dx\) which has the indefinate form: \(\int 2x-9dx=x^2-9x\). This allows me to solve for the area under the curve:\((8.5)^2-9(8.5)-[(4.5)^2-9(4.5)]=\boxed{16}\)
4
Find the area of the region bounded by the graphs of the given equations:
\(y=x^2-2x-2\) and \(y = x+2\).
Answer:
The area will be the difference between the integrals of each such that \(\text{Area} = \int_{-1}^{4}x+2 dx- \int_{-1}^{4}x^2-2x-2 dx\) which leads to:
\(\frac{x^2}{2}+2x + c - \frac{x^3}{3}-x^2-2x+c\). When evaluated over the range \([-1,4]\), the result is: \(\boxed{\frac{125}{6}\approx 20.833}\)
Here is a small plot of the bounded area described:
curve(x^2 -2*x-2, -1, 4)
curve(x+2, add = TRUE, -1,4)
5
A beauty supply store expects to sell \(110\) flat irons during the next year. It costs \(\$3.75\) to store one flat iron for one year. There is a fixed cost of \(\$8.25\) for each order. Find the lot size and the number of orders per year that will minimize inventory costs.
Answer:
Let \(c\) be the cost, \(x\) bet the number of orders in a year and \(i\) the quantity of irons per order. This yeilds:\(c = \$8.25 \cdot x + \frac{\$3.75 \cdot i}{2}\). and since \(x \cdot i = \$110 \implies i = \frac{\$110}{x}\) if I sub in I get: \(c = \$8.25 \cdot x + \frac{\$206.25}{i}\). To maximize, I differentiate on \(i\) to get: \(\frac{d}{dx}c = \$8.25 -\frac{\$206.25}{i}\), setting \(\frac{d}{dx}c =0\) and solving to get: \(i = \sqrt{\frac{\$206.25}{\$8.25}} \boxed{\approx \text{5 per year.}}\)
6
Use integration by parts to solve \(\int \ln(9x)\cdot x^6 dx\).
Answer:
\(\int\ln(9x) \cdot x^6 dx\) and then setting \(u = \ln(9x)\) and \(dv = x^6dx\) I get: \(v = \frac{1}{7}x^7\) and $du = dx $. Then, using the form:
\[\int udv = uv - \int v du \]
I get: \(\ln(9x) x^7 -\frac{1}{7}\int x^6dx = \boxed{x^7\ln(9x)-\frac{x^7}{49}}\).
7
Determine wheather \(f(x)\) is a probability density function on the interval \([1, e^6]\). If not, determine the value of the definite integral: \(f(x)=\frac{1}{6x}\).
Answer:
\[F(x) = \int_1^{e^{6}}\frac{1}{6x}dx=\dots= \frac{1}{6}\big[{\ln(e^6) - \ln(1)}\big]=1\]
\(\therefore\) \(f(x)\) is a prob func. on the interval \([1, e^6]\).